This is a very typical (text-book) example of an active Twin-T notch filter with adjustable Q.

enter image description here

However, after having posted THIS question, one of the commenters pointed out that there is a design that completely leaves out the OpAmp (OA) indicated, as it doesn't seem to serve a clear function/purpose.

I'm currently working on a design involving a high-precision Instrument Amplifier (IA), that is chained through a bunch of filters. The first one being the 60 Hz notch filter shown above, which is then output to an active 100 Hz LPF. The LPF consist of a typical 2nd Order Sallen-Key LPF, followed by a passive 1st Order RC LPF.

All the designs I have seen, are basically using this text-book Twin-T design with 1 or 2 OA's, for output and low impedance feedback, respectively. But are all these designs just a copy-paste artifact of people not thinking by themselves or are there important reasons for having the (indicated) OA in the circuit?

I have simulated the above (original) circuit here:

enter image description here

and and then again the single OA solution here:

enter image description here

This design show a pretty good response of -30dB @ 61.4 Hz (slightly off from original) with -3dB at ~65.6 Hz when using an 11kΩ pot set at 381Ω. But the original design seem to have better accuracy for \$f_c\$.

Q: What are the effects of removing the OA from the circuit?
(What side effects will this have on i/o impedance and component value sensitivity etc? Other pros/cons?)

  • 2
    \$\begingroup\$ The complete analysis of this filter is proposed page 341 in amazon.com/Linear-Circuit-Transfer-Functions-Introduction/dp/…. Getting rid of the op-amp will bring the load and the pot-meter values in the time constant expressions, probably affecting the selectivity at the end and bringing-in unwanted variability. \$\endgroup\$ – Verbal Kint Apr 20 '19 at 15:24
  • \$\begingroup\$ Thanks for that feedback (!), but I don't have that book, or I'd be happy to read all about it there. \$\endgroup\$ – not2qubit Apr 21 '19 at 0:38
  • \$\begingroup\$ So at the end of the day, it looks as though I'll keep both my ops. \$\endgroup\$ – not2qubit Apr 21 '19 at 0:41

Both opamps in the original circuit are wired as voltage followers — unity gain buffers. They have very high input impedance and very low output impedance, which means that they isolate the circuit on the input from any loading by the circuit connected to the output.

As you have discovered, this means that the input circuit performs closer to its theoretical predictions, and you have more freedom to use the output signal any way you wish.

Removing those opamps means that everything now depends on everything else, making prediction of the circuit performance that much more difficult. You CAN make the circuit perform the same function as the circuit with the opamps, but there will be undesirable interactions between circuit parameters, input source impedance and load impedance.

  • \$\begingroup\$ Well said. OP never mentioned the input impedance of the next stage. \$\endgroup\$ – Randy Nuss Apr 20 '19 at 16:18
  • \$\begingroup\$ @RandyNuss The output is low impedance in both cases. It comes directly from an op-amp output. \$\endgroup\$ – Spehro Pefhany Apr 20 '19 at 16:32
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    \$\begingroup\$ @SpehroPefhany I see a low impedance output in both solutions however if the Why This? amplifier is removed in solution 1, the output won't be isolated. I upvoted you. \$\endgroup\$ – Randy Nuss Apr 20 '19 at 23:24

In the 2nd circuit the DC gain varies (with the pot setting) from 1 to the open loop gain of the amplifier. In the first it is always 1 and the pot varies Q.

Not sure how you would like us to compare two circuits that have such different behavior.

  • \$\begingroup\$ From a non EE expert point of view, and apart the already mentioned impedance, they look and behave in an extremely similar way. Perhaps you can expand further on the differences? \$\endgroup\$ – not2qubit Apr 21 '19 at 9:29
  • \$\begingroup\$ As far as I'm concerned the pot-adjustable Q is the key feature in the original circuit. The second circuit behaves completely differently with respect to pot rotation. If you have a fixed Q you don't need two op-amps. \$\endgroup\$ – Spehro Pefhany Apr 21 '19 at 14:54

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